Question

Using integration, find the area of the region bounded by the triangle whose vertices are (2, 1), (3, 4) and (5, 2).

Answer 1

\[\text{ Consider the points A(2, 1), B(3, 4) and C(5, 2) }\]
\[\text{ We need to find area of shaded triangle ABC }\]
\[\text{ Equation of AB is }\]
\[y - 1 = \left( \frac{4 - 3}{3 - 2} \right)\left( x - 2 \right)\]
\[ \Rightarrow 3x - y - 5 = 0 . . . \left( 1 \right)\]
Equation of BC is
\[y - 4 = \left( \frac{2 - 4}{5 - 3} \right)\left( x - 3 \right)\]
\[ \Rightarrow x = y - 7 = 0 . . . \left( 2 \right)\]
Equation of CA is
\[ y - 2 = \left( \frac{2 - 1}{5 - 2} \right)\left( x - 5 \right)\]
\[ \Rightarrow x - 3y + 2 = 0 . . . \left( 3 \right)\]
\[\text{ Area of }\Delta ABC = \text{ Area of }\Delta ABD +\text{ Area of }\Delta DBC\]
\[\text{ In }\Delta ABD, \]
\[\text{ Consider point }P(x, y_2 ) \text{ on AB and }Q(x, y_1 )\text{ on AD }\]
\[\text{ Thus, the area of approximating rectangle with length }= \left| y_2 - y_1 \right| \text{ and width }= dx\text{ is }\left| y_2 - y_1 \right| dx\]
\[\text{ The approximating rectangle moves from }x = 2\text{ to }x = 3\]
\[ \therefore\text{ Area of }\Delta ABD = \int_2^3 \left| y_2 - y_1 \right| dx = \int_2^3 \left( y_2 - y_1 \right) dx \]
\[ \Rightarrow A = \int_2^3 \left( \left( 3x - 5 \right) - \left( \frac{x + 1}{3} \right) \right) dx\]
\[ \Rightarrow A = \int_2^3 \frac{\left( 9x - 15 - x - 1 \right)}{3} dx \]
\[ \Rightarrow A = \int_2^3 \frac{\left( 8x - 16 \right)}{3} dx\]
\[ \Rightarrow A = \frac{1}{3} \left[ 8\frac{x^2}{2} - 16x \right]_2^3 \]
\[ \Rightarrow A = \frac{1}{3}\left[ \left( 4 \times 3^2 \right) - \left( 16 \times 3 \right) - \left( 4 \times 2^2 \right) + \left( 16 \times 2 \right) \right]\]
\[ \Rightarrow A = \frac{1}{3}\left( 68 - 64 \right)\]
\[ \Rightarrow A = \frac{4}{3}\text{ sq . units }\]
\[\text{ Similarly, for }S(x, y_4 )\text{ on AB and }R(x, y_3 )\text{ on DC }\]
\[\text{ Area of approximating rectangle of length }\left| y_4 - y_3 \right|\text{ and width }dx = \left| y_4 - y_3 \right| dx\]
\[\text{ Approximating rectangle moves from }x = 3\text{ to }x = 5\]
\[ \therefore\text{ Area BDC }= \int_3^5 \left| y_4 - y_3 \right| dx\]
\[ \Rightarrow A = \int_3^5 \left( \left( 7 - x \right) - \frac{\left( x + 1 \right)}{3} \right) dx\]
\[ \Rightarrow A = \frac{1}{3} \int_3^5 \left( 20 - 4x \right) dx\]
\[ \Rightarrow A = \frac{1}{3} \left[ 20 x - 4\frac{x^2}{2} \right]_3^5 \]
\[ \Rightarrow A = \frac{1}{3}\left[ \left( 100 - 50 \right) - \left( 60 - 18 \right) \right]\]
\[ \Rightarrow A = \frac{1}{3}\left( 50 - 42 \right) = \frac{8}{3}\text{ sq . units }\]
\[ \therefore\text{ Area of } \Delta ABC =\text{ Area of }\Delta ABD +\text{ Area of }\Delta DBC = \frac{4}{3} + \frac{8}{3} = \frac{12}{3} = 4 \text{ sq . units }\]